690. Employee Importance
Description
You are given a data structure of employee information, which includes the employee's unique id, his importance value and his direct subordinates' id.
For example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. They have importance value 15, 10 and 5, respectively. Then employee 1 has a data structure like [1, 15, [2]], and employee 2 has [2, 10, [3]], and employee 3 has [3, 5, []]. Note that although employee 3 is also a subordinate of employee 1, the relationship is not direct.
Now given the employee information of a company, and an employee id, you need to return the total importance value of this employee and all his subordinates.
Example 1:
Input: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1
Output: 11
Explanation:
Employee 1 has importance value 5, and he has two direct subordinates: employee 2 and employee 3. They both have importance value 3. So the total importance value of employee 1 is 5 + 3 + 3 = 11.
Note:
One employee has at most one direct leader and may have several subordinates.
The maximum number of employees won't exceed 2000.
Solution
DFS
遍历List时间上有点慢,可以加一个map来辅助,但是懒得写啦,不写也能AC。
/*
// Employee info
class Employee {
// It's the unique id of each node;
// unique id of this employee
public int id;
// the importance value of this employee
public int importance;
// the id of direct subordinates
public List<Integer> subordinates;
};
*/
class Solution {
public int getImportance(List<Employee> employees, int id) {
Employee target = null;
for (Employee employee : employees) {
if (employee.id == id) {
target = employee;
break;
}
}
int importance = target.importance;
for (int subId : target.subordinates) {
importance += getImportance(employees, subId);
}
return importance;
}
}
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